Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: elt2112 on January 12, 2015, 04:22:49 PM
-
I'm struggling with this question a bit.
So in Rh(PtBu3)2(H)(CO) the rhodium is in the +1 oxidation state and has 8 valence electrons. So thought that after reacting with C2F2 it would form Rh(PtBu3)2(F)(CO) as this has the same metal electron count, and there is no hydride. I'm a bit stuck with the next bit and can't seem to find anything about it to help.